Laplacian of a 2-form


by haushofer
Tags: 2form, laplacian
haushofer
haushofer is offline
#1
May14-10, 09:59 AM
Sci Advisor
P: 869
Hi,

According to eg Nakahara's conventions the Laplacian on a form K is given by

[tex]
\Delta K = (dd^{\dagger} + d^{\dagger}d)K
[/tex]

In my case K is a two form living in R^3. I've calculated the Laplacian and arrive at

[tex]
\Delta K = \Bigl( \frac{1}{3!}\epsilon^{klm}\epsilon^n_{\ ij}\partial_k \partial_n K_{lm} - \frac{1}{4}\partial_{i}\partial^k K_{jk} \Bigr) dx^i \wedge dx^j
[/tex]

However, the answer seems a little odd to me. Does anyone have a reference where Laplacians of two forms are evaluated, or some comment? Thanks in forward! :)
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haushofer
haushofer is offline
#2
May14-10, 02:36 PM
Sci Advisor
P: 869
And another question: is on a compact manifold a harmonic form neccesarily constant, or is this only true for 0-forms (functions)?
zhentil
zhentil is offline
#3
May17-10, 03:24 PM
P: 491
How would you go about defining a constant form?

The obvious answer to this would be that you can't define it in a coordinate-invariant way. The generalization of the idea of constant functions to forms is with the definition: a constant function is d and d* closed.

st0ck
st0ck is offline
#4
May27-10, 08:58 AM
P: 3

Laplacian of a 2-form


For your first question, if the codifferential is defined with a sign (-1)^n(p+1), it can be shown that in euclidean space in general laplace beltrami of a form is the same as the usual old laplacian of every function factor (times the wedged covectors). It is a lengthy calculation and took me some hours, but it works ;)
haushofer
haushofer is offline
#5
May28-10, 03:36 PM
Sci Advisor
P: 869
Hi,

but that would mean that my calculation isn't right?

About this "constant form": that should be zero, ofcourse; if a tensor is zero in one coordinate system, it's zero in every coordinate system. My bad ;)


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